Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+6y &= 6 \\ -4x+8y &= -2\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $8y = 4x-2$ Divide both sides by $8$ to isolate $y$ $y = {\dfrac{1}{2}x - \dfrac{1}{4}}$ Substitute this expression for $y$ in the first equation. $-8x+6({\dfrac{1}{2}x - \dfrac{1}{4}}) = 6$ $-8x + 3x - \dfrac{3}{2} = 6$ Simplify by combining terms, then solve for $x$ $-5x - \dfrac{3}{2} = 6$ $-5x = \dfrac{15}{2}$ $x = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $x$ back into the top equation. $-8( -\dfrac{3}{2})+6y = 6$ $12+6y = 6$ $6y = -6$ $y = -1$ The solution is $\enspace x = -\dfrac{3}{2}, \enspace y = -1$.